Problem: Let $h(x)=\sqrt{4x-8}$ and let $c$ be the number that satisfies the Mean Value Theorem for $h$ on the interval $[3,11]$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $3.5$ (Choice B) B $6$ (Choice C) C $8$ (Choice D) D $9.5$
Explanation: According to the Mean Value Theorem, there exists a number $c$ in the open interval $(3,11)$ such that $h'(c)$ is equal to the average rate of change of $h$ over the interval: $h'(c)=\dfrac{h(11)-h(3)}{(11)-(3)}$ First, let's find that average rate of change: $\dfrac{h(11)-h(3)}{(11)-(3)}=\dfrac{6-2}{8}={\dfrac{1}{2}}$ Now, let's differentiate $h$ and find the $x$ -value for which $h'(x)={\dfrac{1}{2}}$. $h'(x)=\dfrac{1}{\sqrt{x-2}}$ The solution of $h'(x)=\dfrac{1}{2}$ is $x=6$. $x=6$ is indeed within the interval $(3,11)$. In conclusion, $c=6$.